Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a clever way to determine a primitive element of the finite extension $$F=\mathbb{Q}(\sqrt{2}+i,\sqrt{3}-i)/\mathbb{Q} \text{ ?}$$ On simpler examples, I've been able to find one by determining all field morphisms $\sigma: F\to\mathbb{C}$ such that $\sigma|_\mathbb{Q}=\text{id}$ and finding $x\in F$ with a different image under each such $\sigma$. But here, it seems quite painful since the degree is between 8 and 16...

share|improve this question
4  
The degree is at most $8$, since the field is contained in $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$, which has degree $8$ over $\mathbb{Q}$. –  Arturo Magidin Jun 20 '12 at 18:21

2 Answers 2

The Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$ over $\mathbb{Q}$ is of order $8$; it should not be hard to verify that it is in fact isomorphic to $C_2\times C_2\times C_2$; the generators are complex conjugation, the map that sends $\sqrt{2}$ to $-\sqrt{2}$ and fixes $\sqrt{3}$ and $i$; and the map that sends $\sqrt{3}$ to $-\sqrt{3}$ and fixes $\sqrt{2}$ and $i$.

Now, your field contains $\sqrt{2}+\sqrt{3}+i$. It has eight different images under the automorphisms of $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$: $\pm\sqrt{2}\pm\sqrt{3}\pm i$. So $\mathbb{Q}(\sqrt{2}+\sqrt{3}+i) = \mathbb{Q}(\sqrt{2},\sqrt{3},i)$. Also, your field is contained in $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$, hence in $\mathbb{Q}(\sqrt{2}+\sqrt{3}+i)$.

Your field contains $\sqrt{2}+\sqrt{3}$, and so contains $\sqrt{2}$ and $\sqrt{3}$ (since $\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \mathbb{Q}(\sqrt{2},\sqrt{3})$). So it is equal to $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$, and you are done.

share|improve this answer

It is relatively straightforward to show that $F=\mathbb{Q}(i,\sqrt2,\sqrt3)$, because it is easy to see that $\mathbb{Q}(\sqrt2+i)=\mathbb{Q}(\sqrt2,i)$, and similarly $\mathbb{Q}(\sqrt3-i)=\mathbb{Q}(\sqrt3,i)$. Thus $[F:\mathbb{Q}]=8$, this extension is Galois, and the Galois group is elementary 2-abelian. An automorphism $\sigma$ is uniquely determined once we specify $\sigma(i)=\pm i$ (two choices), $\sigma(\sqrt2)=\pm\sqrt2$ and $\sigma(\sqrt3)=\pm\sqrt3$ (again two choices). If an element $z\in F$ is not fixed by any of these automorphisms, it is not in any of the subfields, so for example $$ z=2i+3\sqrt2+5\sqrt3 $$ will be a primitive element.

share|improve this answer
    
Did you choose these coefficients on $i, \sqrt2, \sqrt3$ for any particular reason? –  Dylan Moreland Jun 20 '12 at 18:45
    
No :-) ${}{}{}{}$ –  Jyrki Lahtonen Jun 20 '12 at 18:46
    
Oh, I didn't remark that $F=\mathbb{Q}(i,\sqrt{2},\sqrt{3})$. Then it's much easier, thanks! –  Klaus Jun 20 '12 at 18:51
    
@Klaus: It is easier than checking that $z$ is not in any of the 14 intermediate fields, and much easier than showing directly that we can write the generators of $F$ as rational expressions of $z$. –  Jyrki Lahtonen Jun 20 '12 at 18:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.